Question

A CYLINDRICAL VESSEL OF 5CM RADIUS IS FILLED WITH WATER UPTO A HEIGHT OF 20CM . THE CYLINDER IS OPEN TO ATMOSPHERE AT THE TOP . A SMALL APERTURE OF RADIUS 2MM IS MADE ON THE SIDE OF CYLINDER AT A HEIGHT OF 5CM FROM THE BOTTOM OF VESSEL . FOR APPROXIMATELY HOW LONG WATER LEAK OUT OF APERTURE

Answer 1

Answer:

Explanation:

P=P0−ρgh=98×103N/m2

P0V0=PV

105[A(500−H)]=98×103[A(500−200)]

H = 206 mm

Level fall = 206 − 200 = 6 mm

Answer 2

Answer:

the time required to leak out if the aperture will be  = 1 minute and 48 seconds.

Explanation:

It is given:

height, H = 20cm - 5cm = 15 cm = 15 x 10⁻² m

Area₁ = A = 25π x 10⁻⁴ m

Area₂ = A₀ = 2π x 10⁻³ m

Considering, g = 10 m/s²

So,. we know that

time , t  = $\frac{A}{A0} \sqrt{\frac{2H}{g} }$

Putting all the values, we get :

t = $\frac{25*10^{-2} }{2*10^{-2} } \sqrt{\frac{2*15*10^{-2}}{10} }$

or, t = 108 s

or, t = 1 minute 48 seconds.

therefore, the time required to leak out if the aperture will be  = 1 minute and 48 seconds.

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